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Question Number 199775 by Mingma last updated on 09/Nov/23

Answered by mathfreak01 last updated on 09/Nov/23

$$$$$$$$Assuming when have a triangle such as this $$DC=\sqrt{{DB}^2+{BC}^2}=\sqrt{5}$$$$AD=DC=\sqrt{5}$$$$\therefore \mathrm{\angle }DAC=\mathrm{\angle }ACD$$$$\mathrm{\angle }DAC+\mathrm{\angle }ACD=\mathrm{\angle }BDC=\theta$$$$\Rightarrow \mathrm{\angle }DAC=\mathrm{\angle }ACD=\frac{\theta }{2}$$$$From△ABC,$$$$\tan \frac{\theta }{2}=\frac{2}{1+\sqrt{5}}$$$$\theta =2\arctan \frac{2}{1+\sqrt{5}}$$$$$$$$From△BDC,$$$$\tan \theta =2$$$$\theta =\arctan 2$$$$$$$$\Rightarrow \arctan 2=2\arctan \frac{2}{1+\sqrt{5}}$$$$$$

Commented by Mingma last updated on 09/Nov/23

Very elegant solution!

Answered by MM42 last updated on 09/Nov/23

$$tan\left(2act\left(\frac{\sqrt{5}-1}{2}\right)\right)=tan\left(act2\right)$$$$\Rightarrow \frac{4(\sqrt{5}-1)}{4-{(\sqrt{5}-1)}^2}=2$$$$\Rightarrow 2(\sqrt{5}-1)=4-5-1+2\sqrt{5}✓$$

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